Working on some exam practice questions and trying to get a better understanding of the solutions.
Suppose that $f$ is an analytic function on an open set $D$ with isolated singularity at $z_o$ and supposed that we have already proven that $|f(z)| \rightarrow \infty$ as $z \rightarrow z_0$
The 2nd part of the question says:
Show that $\frac{1}{f(z)}$ has a removable singularity at $z_0$, and
that there exists $m \ge 1$ and an analytic function $h$ such that
$h(z_0) \ne 0$ and $$\frac{1}{f(z)} = (z-z_0)^mh(z)$$
So the solution says:
Since $|f(z)| \rightarrow \infty$ as $z \rightarrow z_0$, $\exists \space \delta > 0$ such that $|f(z)|> 1$ whenever $0 < |z-z_0|< \delta$. So $\frac{1}{f(z)}$ is well defined and analytic on $B(z_0,\delta) \backslash {z_0}$ and $z_0$ is an isolated singularity.
My question at this point is this: Why does that make $1/f$ analytic? I'm failing to see why that necessarily means we can represent $f(z)$ with a power series for all $ z \in B(z_0,\delta) $?
The solution goes on to say:
Since $$\lim_{z \rightarrow z_0} \frac{1}{f(z)} = 0 \space \space (\star) $$ $z_0$ is an isolated singularity. Why is this the case? Is this because we know $z_0$ is an isolated singularity and since the limit is $0$ we know it's not a pole and not an essential singularity (as the limit exists)?
Therefore it has a power series of the form $\sum_{k=0}^{\infty}a_k (z-z_0)^{k} $ Finally, it says (I don't understand this part), due to $(\star)$, there exists $m \ge 1$ such that $a_k = 0$ for all $k = 1,2,…,m-1$, so therefore: $$\frac{1}{f(z)} = \sum_{k=m}^{\infty}a_k (z-z_0)^{k} = (z-z_0)^{m}\sum_{k=m}^{\infty}a_k (z-z_0)^{k-m} = (z-z_0)^mh(z)$$ Then $h$ is analytic on $B(z_0,\delta)$ and $h(z_0)=a_m \ne 0$.
How do we get from $(\star)$ to the next part? I am failing to see the connection there, could someone help me with explanation?
Many thanks, and any help would be greatly appreciated!
Best Answer
First question: A function is (complex) analytic if and only if it is holomorphic. It is an easy exercise to see that if $f$ is complex differentiable in $w$ and $f(w) \neq 0$, then $1/f$ is also complex differentiable in $w$ and $(1/f)'(w) = -f'(w)/f(w)^2$. So we know that $1/f$ is holomorphic in $B(z_0,\delta)\setminus \{z_0\}$, hence analytic.
Second question: It is not correct that $z_0$ is an isolated singularity of $1/f$ since $\lim\limits_{z\to z_0} \frac{1}{f(z)} = 0$. It is an isolated singularity since $1/f$ is holomorphic in $B(z_0,\delta)\setminus\{z_0\}$. The limit says that $z_0$ is a removable singularity, and the value which removes the singularity is $0$, so
$$g(z) = \begin{cases}\frac{1}{f(z)} &, z \neq z_0\\ 0 &, z = z_0 \end{cases}$$
is holomorphic on $B(z_0,\delta)$.
Third question: Since $g(z_0) = 0$, the power series representation of $g$ about $z_0$ has a constant term zero, hence
$$g(z) = \sum_{k=1}^\infty a_k (z-z_0)^k.$$
since $g$ does not vanish identically, not all coefficients are $0$, and if we let $m = \min \{ k \in\mathbb{N} : a_k \neq 0\}$, we have
$$g(z) = \sum_{k=m}^\infty a_k (z-z_0)^k$$
with $a_m \neq 0$. Then
$$h(z) = \sum_{k=m}^\infty a_k (z-z_0)^{k-m} = \sum_{r=0}^\infty a_{r+m}(z-z_0)^r$$
is holomorphic in $B(z_0,\delta)$ with $h(z_0) = a_m \neq 0$, and evidently $g(z) = (z-z_0)^m\cdot h(z)$. Since $g$ has no zero in $B(z_0,\delta)\setminus\{z_0\}$, $h$ does not vanish anywhere in $B(z_0,\delta)$, so $\tilde{h}(z) = \frac{1}{h(z)}$ is holomorphic on $B(z_0,\delta)$, and we have
$$f(z) = (z-z_0)^{-m}\cdot \tilde{h}(z)$$
on $B(z_0,\delta)\setminus\{z_0\}$.